Acid/Base Properties of Salts Hiding in plain sight

Recognizing Bases Sometimes it seems that all acids and bases are labeled with H + or OH - Acids do have an H + But a base is ANYTHING that can accept a proton – it doesn’t have to have an OH-. Anything with a negative charge or just non-bonding electrons (N, O, P) can act as a base. The salt formed by the dissociation of an acid is the “conjugate base” of the acid. This must mean that the salt is a base, whether it has an OH- or not.

Pick a salt, any salt  How about ammonium acetate, NH 4 OAc? An excellent choice.  SALT = IONIC  Ammonium acetate is an ionic solid.  Ionic solids dissociate in aqueous solution  Aqueous NH 4 OAc will exist as anions and cations.

The MOST important thing about salts! NH 4 OAc (aq) NH 4 + (aq) + OAc - (aq) So what can we say about NH 4 + and OAc - in aqueous solution? We need to think “backwards”. How did NH 4 OAc get the NH 4 + in the first place? (Or, more accurately, what is one way it could have gotten it?)

Salts are products of acid/base reactions You might recall that the reaction of an acid and a base yields a salt and water: NH 4 OH + HOAc → H 2 O + NH 4 OAc NH 3 + HOAc  NH 4 OAc These are the same reaction! Hey, look! It’s our salt!

These two reactions are identical NH 3 + H 2 O ↔ NH OH - K b (NH 3 ) NH 3 + H 2 O = NH 4 OH NH 4 OH + HOAc → H 2 O + NH 4 OAc NH 3 + H 2 O + HOAc → H 2 O + NH 4 OAc NH 3 + HOAc  NH 4 OAc This is why I often say that bases don’t need to have an OH- because they borrow one from water. The whole acid + base = salt + water is a slight exaggeration. Sometimes you borrow OH- from water! NH 4 OH is just another way to write aqueous NH 3

A more detailed reaction Let’s stick to the more plain: NH 3 (aq) + HOAc → NH 4 OAc(aq) But what are NH 3 (aq) and HOAc (aq) ?

Acid (and Base) Dissociation Reactions NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq) K b HOAc (aq) + H 2 O (l)  H 3 O + (aq) + OAc - (aq) K a Hey, look! It’s pieces of our salt!

Putting it all together… NH 4 OAc (aq) NH 4 + (aq) + OAc - (aq) NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq) K b Baseacid conjugate conjugate acid base HOAc (aq) + H 2 O (l)  H 3 O + (aq) + OAc - (aq) K a Acidbase conjugate conjugate acid base

In short… We have the conjugate acid (NH 4 + ) of a base (NH 3 or NH 4 OH), and the conjugate base (OAc - ) of an acid (HOAc). THIS IS A GENERAL RULE! All salts are the combination of a (conjugate) acid and a (conjugate) base!

An acid is an acid. A base is a base. “Conjugate” or not. And water is both!

K b becomes K a as K a becomes K b Just because you’re a “conjugate acid” doesn’t mean you’re not an acid! NH 4 + (aq) + H 2 O (l)  NH 3 (aq) + H 3 O + (aq) K a =K w /K b OAc - (aq) + H 2 O (l)  OH - (aq) + HOAc (aq) K b =K w /K a

But one is strong and the other weak.

Important Points  All acids/bases are conjugate pairs of compounds. It’s an equilibrium, they go both ways.  Conjugate reaction pairs have K a xK b =K w  For this particular salt it is approximately equally “acidic” and “basic” and so it ends up being neutral. This is not usually the case. The K a and K b just happen to be really close in this case.  In general, you’d need to solve both ICE charts to get the pH…but not always.

Pick a new salt, any salt  How about sodium acetate, NaOAc? An excellent choice.  Sodium acetate is an ionic solid.  Ionic solids dissociate in aqueous solution  Aqueous NaOAc will exist as anions and cations.

Salts dissociate in water NaOAc (aq) = Na + (aq) + OAc - (aq) It’s a little less obvious what’s going on here. But ALL SALTS are made up of an acid (cation) part and a base (anion) part. It’s just that Na + doesn’t look like much of an acid. But again, that’s because it borrows from water!

Acid (and Base) Dissociation Reactions NaOH (aq) + H 2 O (l)  H-OH (l) + OH - (aq) + Na + (aq) K b Or: NaOH  Na + (aq) + OH - (aq) OH - (aq) + H 2 O (l)  H-OH (l) + OH - (aq) K b HOAc (aq) + H 2 O (l)  H 3 O + (aq) + OAc - (aq) K a

Putting it all together… NaOAc (aq) Na + (aq) + OAc - (aq) NaOH (aq) + H 2 O (l)  OH - (aq) + Na + (aq) + H-OH (l) HOAc (aq) + H 2 O (l)  H 3 O + (aq) + OAc - (aq)

Putting it all together… NaOAc (aq)  Na + (aq) + OAc - (aq) NaOH (aq) + H 2 O (l)  OH - (aq) + Na + (aq) + H-OH (l) Baseacid conjugate conjugate base acid HOAc (aq) + H 2 O (l)  H 3 O + (aq) + OAc - (aq) Baseacid conjugate conjugate acid base

In short… We have the conjugate acid (Na + ) of a base (NaOH), and the conjugate base (OAc - ) of an acid (HOAc).

K b becomes K a as K a becomes K b NaOAc (aq)  Na + (aq) + OAc - (aq) Na + (aq) + 2 H 2 O (l)  NaOH (aq) + H 3 O + (aq) (or, if you prefer) Na + (aq) + H 2 O (l)  NaOH (aq) + H + (aq) K a =K w /K b OAc - (aq) + H 2 O (l)  OH - (aq) + HOAc (aq) K b =K w /K a

But one is strong and the other weak.

Net Result NaOAc gives rise to a single equilibrium reaction that must be considered: OAc - (aq) + H 2 O (l)  OH - (aq) + HOAc (aq) K b =5.56x The salt is a base!!!!

What is the pH of a M NaOAc solution? We need to recognize this as a salt solution. As soon as we recognize it as a salt, it has an acid half and a base half. In this case, we can ignore the Na + because it comes from a strong base (it ain’t going back!) and so it is only the OAc - that matters.

What is the pH of a M NaOAc solution? It’s a base equilibrium so…THERE’S 3 PARTS!

ICE-ICE-BABY-ICE-ICE OAc - (aq) + H 2 O (l)  OH - (aq) + HOAc (aq) I0.100 M-00 C-x-+x E0.100-x-xx

How do we solve it?

ICE-ICE-BABY-ICE-ICE I0.100 M-00 C-x-+x E

Sample Problem What is the pH of a solution of 15 g/L of potassium phosphate in water?

1 st we need…

…a balanced equation K 3 PO 4 (aq)  3 K + (aq) + PO 4 3- (aq) Now, potassium is the conjugate acid of KOH – a strong base. PO 4 3- is the conjugate base of hydrogen phosphate, HPO 4 2- (which is the conjugate base of H 2 PO 4 -, which is the conjugate base of H 3 PO 4 )

So, there may be 3 equilibria to consider H 3 PO 4 (aq) + H 2 O (l)  H 2 PO 4 - (aq) + H 3 O + (aq) K a1 = 7.5x10 -3 H 2 PO 4 - (aq) + H 2 O (l)  HPO 4 2- (aq) + H 3 O + (aq) K a2 = 6.2x10 -8 HPO 4 2- (aq) + H 2 O (l)  PO 4 3- (aq) + H 3 O + (aq) K a3 = 5.8x But, of course, we need the reverse reactions

So, there may be 3 equilibria to consider

ICE ICE ICE But first, we need suitable units. 15 g/L is an acceptable unit of concentration, BUT an ICE chart requires Molarity. Why?

MOLES! MOLES! MOLES! An ICE chart is simply an accounting trick for a balanced chemical equation. Chemical equations are all about molar relationships, so you can only use moles or Molarity (moles/L)

Simple Conversion 15 g K 3 PO 4 1 mol K 3 PO 4 = M L solution g K 3 PO 4 And now we are ready for some ICE

Just take them 1 at a time… PO 4 3- (aq) + H 2 O (l)  HPO 4 2- (aq) + OH - (aq) I C E M-00 -x-+x x-xx

The 1 st equilibrium… PO 4 3- (aq) + H 2 O (l)  HPO 4 2- (aq) + OH - (aq) I C E M x x x10 -2

…leads to the second HPO 4 2- (aq) + H 2 O (l)  H 2 PO 4 - (aq) + OH - (aq) I C E 2.73x10 -2 M x x -+x+x +x+x 2.73x x - x2.73x x

…leads to the second HPO 4 2- (aq) + H 2 O (l)  H 2 PO 4 - (aq) + OH - (aq) I C E 2.73x10 -2 M x x x x x x10 -2

If we needed a 3 rd chart…it starts there: H 2 PO - (aq) + H 2 O (l)  H 3 PO 4 (aq) + OH - (aq) I C E 1.61x10 -7 M x x -+x+x +x+x 1.61x x - x2.73x x

But if the 2 nd one didn’t matter… …the 3 rd one matters even less…

Calculating the pH The result of our 2 nd ICE chart is that: [OH-]= 2.73x10 -2 What do we do with this number?

Calculating the pH The result of our ICE chart (1 st or 2 nd ) is that: [OH-]= 2.73x10 -2 What do we do with this number? 1. To calculate the pH directly, we need [H 3 O + ] 2. We can calculate the pOH, and then get the PH from that.

Calculating the pH from [OH - ] 1. To calculate the pH directly, we need [H 3 O + ] Recall that K w =1x = [H 3 O + ][OH-] 1x = [H 3 O + ][2.73x10 -2 ] [H 3 O + ] = 3.66x pH = - log ([H 3 O + ]) pH = - log (3.66x ) pH = We can calculate the pOH, and then get the PH from that. Recall that pOH + pH = 14 pOH = - log ([OH-]) = - log (2.73x10 -2 ) = 1.56 pH = 14 – pOH = 14 – 1.56 = 12.44

A few concluding observations  The pH ended up being 12.44, nicely basic as we expected.  This is nothing new: K IS K IS K IS K  While it is a multiple equilibrium problem, as we saw for the polyprotic acids, if there is a significant difference in K, not all of the equilibria will contribute meaningfully.

For all salts IN WATER:  Separate into ions.  The cation is the conjugate acid.  The anion is the conjugate base.  For the cation: either add OH- or, if there is one, remove the H + to see what base “it came from”.  For the anion: add an H + to see what acid “it came from”.  Look up the K a and K b of the “parent” acid/base.  Ignore strong acids and bases!

Consider M NaCl Is it acid, basic or neutral? Separate into ions. NaCl=Na + + Cl - The cation (Na+) is the conjugate acid. The anion (Cl-) is the conjugate base. A. Acid B. Base C. Neutral D. Both an acid and a base E. Your mother

Consider M NaCl Is it acid, basic or neutral?  For the cation: either add OH- or, if there is one, remove the H + to see what base “it came from”. Na + + OH - = NaOH For the anion: add an H + to see what acid “it came from”. Cl - + H + = HCl

Consider M NaCl Is it acid, basic or neutral? Look up the K a and K b of the “parent” acid/base. NaOH – strong base HCl – strong acid We can ignore both of them so the salt is neutral!

Consider M NH 4 Cl Is it acid, basic or neutral? Separate into ions. NH 4 Cl=NH Cl - The cation (NH 4 +) is the conjugate acid. The anion (Cl-) is the conjugate base. A. Acid B. Neutral C. Basic D. None of the Above E. All of the above

Consider M NH 4 Cl Is it acid, basic or neutral?  For the cation: either add OH- or, if there is one, remove the H + to see what base “it came from”. NH H + = NH 3 For the anion: add an H + to see what acid “it came from”. Cl - + H + = HCl

Consider M NH 4 Cl Is it acid, basic or neutral? Look up the K a and K b of the “parent” acid/base. NH 3 – K b =1.76x10 -5 HCl – strong acid We ignore the HCl. The NH 3 counts. Clickers!

What is the pH of M NH 4 Cl A B C D E F G H I. None of the above